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Solution to why that nonic is solvable

Previously, we claimed that for any quadruple of rational integers, not all zero, the nonic polynomial

is solvable, meaning that it is possible to determine the roots of explicitly by radicals. By Galois theory, this is equivalent to the assertion that the Galois group of the Galois closure of is a solvable group.

To do this, we need the following fact, which was proved by Bhargava and Yang in this paper as Theorem 4. The statement of their theorem is correct, and the proof is mostly correct, but there is a minor issue. The problem is that the stabilizer of under , which we will denote by , need not be realizable as a subgroup of the Galois group of , which we will denote by . However, the argument they gave for the commuting action between elements of and is correct.

We now consider a binary form of the shape given above. We see that both and act on the roots of and can therefore be embedded via their action on the roots of into , the symmetric group on nine letters. If we restrict to action and denote by , then it follows from Galois theory that must be a subgroup of the centralizer of any element in in .

We then check that, miraculously, always contains the following element of order 3:

We check that the only complex numbers fixed by this element are the roots of . Therefore, if is irreducible, then no root of can be fixed by . Relabelling the roots if necessary, we can assume that can be realized in as

The centralizer of is the stabilizer of under the action by conjugation of on itself. The orbit of is precisely the set of elements in of the same cycle type, therefore the orbit contains

By the orbit-stabilizer theorem, it follows that contains elements. Since is contained in , it is solvable by Burnside’s theorem, which asserts that any finite group whose order is only divisible by two distinct primes is solvable.